UNIT 2: FLYWHEELS, GOVERNORS, BALANCING, CLUTCHES & BRAKES
1.0 FLYWHEELS AND ENERGY FLUCTUATION
1.1 Function & Application
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Primary Function: To store kinetic energy during the power stroke of an engine and release it during the other strokes, thereby reducing speed fluctuation.
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Distinction from Governor: A flywheel reduces the amplitude of speed fluctuation. A governor controls the mean speed by regulating the energy supply.
1.2 Turning Moment Diagram
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A plot of crank torque (T) vs. crank angle (θ) for one cycle (e.g., 720° for 4-stroke).
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Mean Torque Line (Tₘ): Horizontal line representing constant torque that would do the same net work.
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Interpretation: Areas above Tₘ represent energy surplus (stored in flywheel). Areas below Tₘ represent energy deficit (drawn from flywheel).
Diagram:
DiagramSEARCH: turning moment diagram four stroke engine
1.3 Fluctuation of Energy & Speed
- Fluctuation of Energy (ΔE): The maximum difference between the actual energy at any point and the mean energy. It equals the maximum area of one loop (surplus or deficit) on the turning moment diagram.
$$\Delta E = \text{Maximum area enclosed by torque curve and mean torque line}$$
- Fluctuation of Speed (ΔN): The difference between maximum speed (N₁) and minimum speed (N₂).
$$\Delta N = N_1 - N_2$$
- Coefficient of Fluctuation of Energy (Kₑ): Ratio of ΔE to the work done per cycle (W).
$$K_e = \frac{\Delta E}{W}$$
- Coefficient of Fluctuation of Speed (Kₙ): Ratio of ΔN to the mean speed (Nₘ).
$$K_n = \frac{\Delta N}{N_m}$$
> [!TIP] For many applications, speed fluctuation is limited to **±3%** of mean speed, i.e., $$\displaystyle K_n \approx 0.03 $$.
1.4 Flywheel Design Parameters
- Energy Stored in Flywheel:
$$E = \frac{1}{2} I \omega^2$$
where $$\displaystyle I = m k^2 $$ (Mass moment of inertia), $k$ = radius of gyration, $\omega$ = angular velocity.
- Relation between ΔE and Δω: For small fluctuations,
$$\Delta E \approx I \cdot \omega_m \cdot \Delta \omega$$
where $$\displaystyle \omega_m $$ is mean angular velocity.
- Design Formula: From $$\displaystyle \Delta E = I \omega_m \Delta \omega $$ and $$\displaystyle \Delta \omega = \frac{2\pi \Delta N}{60} $$,
$$I = \frac{\Delta E}{\omega_m \Delta \omega} = \frac{\Delta E \times 60}{2\pi N_m \Delta N}$$
Given $$\displaystyle I = m k^2 $$, mass $m$ or radius $k$ can be found.
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Key Steps for Problem:
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Find net work per cycle (W) from mean torque: $$\displaystyle W = T_m \times \text{cycle angle (rad)} $$.
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Calculate ΔE from diagram areas (convert scale: 1 mm² = (vertical scale × horizontal scale) N-m).
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Apply $$\displaystyle \Delta E = I \omega_m \Delta \omega $$ with $$\displaystyle \Delta N = N_1 - N_2 $$ (given or from $$\displaystyle K_n $$).
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2.0 GOVERNORS
2.1 Function and Classification
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Function: To maintain constant speed of an engine under varying load by automatically adjusting the fuel/steam supply.
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Classification:
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Centrifugal: Watt, Porter, Proell (based on centrifugal force of rotating balls).
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Inertia: Sensitive to inertia forces of the governor itself.
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Simple: One rotating mass/arm set.
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Compound: Two sets of rotating masses (e.g., Porter, Proell).
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2.2 Watt Governor
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Construction: Two balls on arms, hinged to a rotating spindle. Arms connected to a sleeve via links.
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Forces: Weight of ball (W), tension in arm (T), centrifugal force ($$\displaystyle m r \omega^2 $$).
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Equilibrium (Vertical & Horizontal):
$$T \cos \theta = W$$
$$T \sin \theta = m r \omega^2$$
where $r$ = radius of rotation of ball, $\theta$ = angle of arm with vertical.
- Height (h) of Governor:
$$h = \frac{g}{\omega^2}$$
> **Proof:** From above, $$\displaystyle \tan \theta = \frac{m r \omega^2}{W} $$. But $$\displaystyle h = \frac{a}{\tan \theta} $$ (a = distance from hinge to sleeve). For small $\theta$, $r \approx a \theta$, leading to $$\displaystyle h \propto \frac{1}{\omega^2} $$ or $$\displaystyle h \propto \frac{1}{N^2} $$.
\boxed{h = \frac{g}{\omega^2} = \frac{8950}{N^2} \text{ (if h in m, N in rpm)}}
2.3 Porter Governor
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Construction: Similar to Watt, but additional central load (Wₛ) on the sleeve. Upper arms (length l) connect balls to sleeve; lower arms (length l') connect sleeve to spindle.
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Force Analysis: Consider forces on one ball and the sleeve.
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On ball: $$\displaystyle T_1 \cos \theta = W $$, $$\displaystyle T_1 \sin \theta = m r \omega^2 $$.
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On sleeve: $$\displaystyle T_2 \cos \phi = W_s + 2W \cos \theta $$, $$\displaystyle T_2 \sin \phi = 2W \sin \theta $$.
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Geometry: $$\displaystyle \frac{r \sin \theta}{l \cos \theta} = \frac{h \sin \phi}{l' \cos \phi} $$.
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Speed Equation:
$$\omega^2 = \frac{g (W_s + 2W)}{m r} \cdot \frac{l'}{l + l'}$$
> [!TIP] Friction at sleeve **lowers** the effective $$\displaystyle W_s $$, increasing the speed range.
2.4 Proell Governor
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Construction: Lower arms are extended and pivoted on the spindle. Balls attached to these extensions. Upper arms connect to sleeve.
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Key Feature: At minimum speed, the lower arm extensions are parallel to the governor axis.
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Force Analysis: Similar to Porter, but geometry differs. Let lower arm length = $l'$, extension length = $b$.
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Minimum Speed (when $b \parallel$ axis):
$$\omega_{min}^2 = \frac{g (W_s + 2W)}{m r} \cdot \frac{l'}{l + l'}$$
> **Note:** This is **identical** to Porter's equation, but the *effective* $r$ and geometry make Proell more sensitive.
- General Speed Equation: More complex, but for given $r$, solve for $\omega$.
2.5 Governor Performance Characteristics
- Sensitiveness (S):
$$\text{Sensitiveness} = \frac{\text{Change in speed}}{\text{Mean speed}} = \frac{\Delta N}{N_m}$$
Also, $$\displaystyle S = \frac{\text{range}}{\text{mean speed}} $$. A **sensitive** governor has a large $S$ (large speed change for small load change).
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Isochronism: Condition where sensitiveness is infinite ($$\displaystyle \Delta N = 0 $$). Governor maintains constant speed for all loads within its range. Requires controlling force $$\displaystyle F_c \propto r $$.
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Hunting: Oscillations of the governor sleeve about its mean position due to oversensitivity. Caused by too high sensitivity or excessive friction.
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Stability: Governor returns to equilibrium position for any speed change.
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Controlling Force Diagram (F_c vs r):
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Stable: $$\displaystyle F_c $$ curve lies above the centrifugal force line ($$\displaystyle m\omega^2 r $$) for $$\displaystyle r < r_{eq} $$ and below for $$\displaystyle r > r_{eq} $$.
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Unstable: Opposite.
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Isochronous: $$\displaystyle F_c $$ curve coincides with $$\displaystyle m\omega^2 r $$ line.
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Stability Condition: For a stable governor,
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$$\frac{dF_c}{dr} > m\omega^2$$
at the equilibrium radius.
2.6 Comparative Analysis: Porter vs. Proell
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Porter: Lower arms pivot on sleeve. $r$ increases with $\omega$.
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Proell: Lower arms pivot on spindle (extension parallel at min speed). For same $r$, $$\displaystyle \omega_{Proell} < \omega_{Porter} $$.
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Proof of Higher Sensitivity (Proell): For same $$\displaystyle m, W, W_s, l, l' $$, the range of speed (N_max - N_min) is larger for Proell because at minimum speed, the effective lever arm for centrifugal force is longer (due to extension $b$), making the governor respond more to speed change.
2.7 Effects of Friction
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Friction at sleeve opposes motion. It reduces sensitivity.
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Coefficient of Insensitiveness (f): Fraction of load (or speed) range lost due to friction.
$$f = \frac{\text{Loss of range due to friction}}{\text{Total range without friction}}$$
3.0 BALANCING OF ROTATING AND RECIPROCATING MASSES
3.1 Fundamental Concepts
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Primary vs. Secondary Balancing:
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Primary: Balancing forces due to first-order inertia (mass × acceleration of crank).
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Secondary: Balancing forces due to second-order inertia (from connecting rod obliquity, $\cos 2\theta$ term).
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Dynamically Equivalent System: A system of two masses (one at crank radius $r$, another at a distance $b$) that replicates the total mass and mass moment of inertia of the original system.
$$m = m_1 + m_2, \quad m r^2 = m_1 r^2 + m_2 b^2$$
3.2 Balancing of Rotating Masses
- Single Mass in One Plane: Balance by adding a counter-mass in the same plane at opposite angular position.
$$m_b r_b = m r \quad \text{(magnitude and direction opposite)}$$
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Multiple Masses in Different Planes: Use moment and force balance. Choose a reference plane; transfer masses to it using $m \times \text{distance}$.
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Hammer Blow (Locomotives): The unbalanced vertical force due to rotating/reciprocating masses that varies with crank angle. It causes dynamic load on rails.
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For rotating mass only: $$\displaystyle F_v = m \omega^2 r \sin \theta $$
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For reciprocating mass (partially balanced): $$\displaystyle F_v = (1 - c) m \omega^2 r \sin \theta $$ where $c$ = fraction balanced.
Maximum Hammer Blow: $$\displaystyle F_{v,max} = (1-c) m \omega^2 r $$ (when $$\displaystyle \sin\theta = 1 $$).
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3.3 Balancing of Reciprocating Masses
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Inertia Forces:
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Primary: $$\displaystyle F_p = m \omega^2 r \frac{\cos \theta}{\sqrt{1 - (r/l)^2 \sin^2 \theta}} \approx m \omega^2 r \cos \theta $$ (for $$\displaystyle r/l << 1 $$).
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Secondary: $$\displaystyle F_s = m \omega^2 r \frac{r}{l} \cos 2\theta $$ (approx).
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Partial Balancing: Only a fraction (c) of reciprocating mass is balanced by a revolving mass (to avoid excessive vertical force on engine).
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Balanced revolving mass: $$\displaystyle m_b = c \cdot m_r $$ (reciprocating mass $$\displaystyle m_r $$).
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Resultant Unbalanced Force: $$\displaystyle F_{un} = (1-c) m_r \omega^2 r \cos \theta $$ (primary) + secondary terms.
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Swaying Couple (Uncoupled 2-cylinder engine): The couple about the engine's centerline due to unbalanced primary forces when cylinders are at an angle $\alpha$.
$$C = (1-c) m \omega^2 r \cdot d \cdot \cos(\theta - \alpha)$$
where $d$ = distance between cylinder centerlines.
- Variation in Tractive Effort (Locomotives): The net horizontal force on the locomotive frame due to unbalanced reciprocating masses, causing pulsation in pulling force.
$$F_{te} = (1-c) m \omega^2 r \cos \theta \cdot \frac{D}{2d}$$
(for 2-cylinder, $D$ = driving wheel diameter, $d$ = wheelbase).
3.4 Multi-Cylinder Engines
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Radial Engine (e.g., 3-cylinder @ 120°):
- Primary Force:
$$\vec{F}_p = m \omega^2 r \left[ \cos \theta + \cos(\theta+120°) + \cos(\theta+240°) \right] = 0$$
* **Secondary Force:**
$$\vec{F}_s = m \omega^2 r \frac{r}{l} \left[ \cos 2\theta + \cos 2(\theta+120°) + \cos 2(\theta+240°) \right] = 0$$
* **Conclusion:** **Perfectly balanced** for both primary and secondary.
- In-line Engines: Complete balancing of reciprocating masses is impossible without creating large unbalanced couples. Usually, primary forces can be balanced (by arranging phases), but secondary forces and couples remain.
3.5 Balancing Mass Calculations
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For a given configuration (crank angles, masses), calculate resultant unbalanced force vectorially.
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Balancing mass $$\displaystyle m_b $$ placed at radius $$\displaystyle r_b $$:
$$m_b r_b = \text{Resultant unbalance}$$
- Angular position opposite to resultant unbalance vector.
4.0 FRICTION CLUTCHES AND BRAKES
4.1 Clutches
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Single Plate Clutch:
- Torque Transmission:
$$T = \mu p \cdot \text{Mean effective radius} \cdot \text{Area}$$
* **Assumptions:**
1. **Uniform Pressure (p constant):** $$\displaystyle p = \frac{F}{\pi (r_o^2 - r_i^2)} $$
$$T = \mu F \cdot \frac{2}{3} \cdot \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2}$$
2. **Uniform Wear (pr = constant):** $$\displaystyle p = \frac{F r_i}{\pi (r_o^2 - r_i^2) r} $$
$$T = \mu F \cdot \frac{1}{2} (r_o + r_i)$$
* **Mean Radius:**
$$R_m = \frac{2}{3} \cdot \frac{r_o^3 - r_i^3}{r_o^2 - r_i^2} \text{ (U.P.)}, \quad R_m = \frac{r_o + r_i}{2} \text{ (U.W.)}$$
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Conical Clutch:
- Torque:
$$T = \mu F \cdot \frac{r}{\sin \alpha}$$
where $\alpha$ = cone angle, $r$ = mean radius.
* **Power Loss:** $$\displaystyle P_f = T_f \cdot \omega $$, where $$\displaystyle T_f = \mu F \cdot R_m $$ (frictional torque).
4.2 Brakes
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Band Brake (with lever):
- Tension Ratio:
$$\frac{T_1}{T_2} = e^{\mu \theta}$$
(where $\theta$ in radians, $$\displaystyle T_1 > T_2 $$).
* **Braking Torque:**
$$T_b = (T_1 - T_2) \cdot r_d$$
* **Lever Analysis:** Use moments about fulcrum to find $$\displaystyle T_1 $$ and $$\displaystyle T_2 $$ from effort $$\displaystyle F_e $$.
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Internal Expanding Shoe Brake:
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Self-energizing: Leading shoe gets additional force from friction on its leading edge.
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Force Analysis: For each shoe, resolve forces radially and tangentially at shoe tip. Sum moments about pivot to find required force $F$ for given $$\displaystyle T_b $$.
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Width (b): From pressure limit $$\displaystyle p_{max} $$:
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$$F = p_{avg} \cdot (b \cdot \text{arc length})$$
where $$\displaystyle p_{avg} = \frac{p_{max}}{2} $$ (approx for uniform wear).
- Double Shoe Brake: Symmetrical; spring force $$\displaystyle F_s $$ determined from torque equation for one shoe (considering both shoes share load).
5.0 FRICTION BEARINGS
5.1 Conical Pivot Bearing
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Geometry: Cone angle = $2\alpha$, shaft radius = $r$, outer radius = $R$, inner radius = $r$ (if hollow).
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Uniform Pressure Assumption:
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Pressure $$\displaystyle p = \frac{W}{\pi (R^2 - r^2)} $$ (constant).
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Frictional Torque:
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$$T_f = \frac{2}{3} \mu W \cdot \frac{R^3 - r^3}{R^2 - r^2}$$
* **Power Loss:** $$\displaystyle P_f = T_f \cdot \omega $$.
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Uniform Wear Assumption (pr = constant):
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Pressure $$\displaystyle p = \frac{W}{2\pi (R^2 - r^2) r} $$ (varies as $1/r$).
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Frictional Torque:
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$$T_f = \frac{1}{2} \mu W (R + r)$$
> [!TIP] Uniform wear gives **lower torque** than uniform pressure for same $W$.
5.2 Collar Bearing
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Geometry: Shaft diameter $d$, collar outer diameter $D$, number of collars $n$.
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Uniform Pressure Assumption:
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Pressure $$\displaystyle p = \frac{W}{n \cdot \frac{\pi}{4} (D^2 - d^2)} $$.
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Frictional Torque:
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$$T_f = \frac{\mu W n}{4} \cdot \frac{D^3 - d^3}{D^2 - d^2}$$
* **Power Absorbed:** $$\displaystyle P_f = T_f \cdot \omega $$.
* **Number of Collars:** From $$\displaystyle p \leq p_{allow} $$,
$$n \geq \frac{4W}{\pi p_{allow} (D^2 - d^2)}$$
6.0 DYNAMOMETERS
6.1 Classification
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Absorption Dynamometers: Absorb engine power as heat (Prony brake, Rope brake).
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Transmission Dynamometers: Transmit power while measuring (Epicyclic, Belt).
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Torsion Dynamometers: Measure torque via shaft angle of twist.
6.2 Torsion Dynamometer
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Construction: Shaft with strain gauges or opticalencoder at two points separated by known length $L$.
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Working: Torque $T$ causes angle of twist $\phi$:
$$\phi = \frac{T L}{G J}$$
where $G$ = shear modulus, $J$ = polar moment of inertia.
- Power Calculation:
$$P = T \cdot \omega = \left( \frac{G J \phi}{L} \right) \cdot \omega$$
$\omega$ in rad/s, $P$ in Watts.
7.0 KINEMATICS OF MECHANISMS
7.1 Four-Bar Chain Analysis
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Velocity (Instantaneous Center Method):
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Locate all ICs (including at infinity for parallel velocities).
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Use $$\displaystyle v = \omega \cdot r $$ for links with known $\omega$.
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Relative velocity: $$\displaystyle v_{B/C} = v_B - v_C $$.
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Acceleration:
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Equation: $$\displaystyle \vec{a}_B = \vec{a}_A + \vec{a}_{B/A}^{rel} + \vec{a}_{B/A}^{cor} $$
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Coriolis Component: $$\displaystyle a_c = 2 \omega \cdot v_{rel} $$ (perpendicular to relative velocity).
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Numerical: Given link lengths, $$\displaystyle \omega_{AB} $$, $$\displaystyle \alpha_{AB} $$, find $$\displaystyle \omega_{BC}, \alpha_{BC}, \omega_{CD}, \alpha_{CD} $$ at a specific configuration.
7.2 Cam and Follower Dynamics (Offset Cam, Flat Faced Follower)
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Offset Cam: Cam center $O$ offset by $e$ from camshaft axis $A$. Follower line of action through $A$.
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Follower Lift (y): For a circular cam of radius $R$ rotating at $\omega$,
$$y = R - \sqrt{R^2 - e^2 \sin^2 \omega t} - e \cos \omega t$$
- Follower Acceleration (a): Differentiate twice w.r.t. time.
$$a = \frac{d^2 y}{dt^2} = \omega^2 \left[ \frac{e^2 R \sin^2 \omega t}{(R^2 - e^2 \sin^2 \omega t)^{3/2}} + e \cos \omega t \right]$$
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Lift-off Condition: When spring force $$\displaystyle F_s = k(y - y_0) $$ becomes zero (or negative). Critical speed $$\displaystyle \omega_c $$ occurs when maximum downward acceleration of follower equals g (if mass is considered) or when spring force just becomes zero at maximum lift.
For given spring stiffness $k$, initial compression $$\displaystyle y_0 $$, find $\omega$ where $$\displaystyle k(y - y_0) = 0 $$ at max $y$.
8.0 MISCELLANEOUS CONCEPTS
8.1 Friction Circle (Journal Bearing)
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Definition: A circle of radius $$\displaystyle r_f $$ representing the locus of the resultant friction force vector as the journal rotates under load.
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Radius:
$$r_f = r \sin \phi$$
where $r$ = journal radius, $\phi$ = angle of friction ($$\displaystyle \tan \phi = \mu $$).
- Significance: Frictional torque $$\displaystyle T_f = W \cdot r_f $$, where $W$ is radial load.
8.2 Dynamically Equivalent System
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Definition: A two-mass system that has the same total mass (m) and same mass moment of inertia (I) as the original system.
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Criteria: For original mass $m$ at distance $r$, equivalent masses $$\displaystyle m_1 $$ (at $r$) and $$\displaystyle m_2 $$ (at $b$) must satisfy:
$$m = m_1 + m_2, \quad m r^2 = m_1 r^2 + m_2 b^2$$
Often $b$ is chosen conveniently (e.g., $$\displaystyle b = 2r $$ or $$\displaystyle b = \infty $$).
8.3 Key Definitions (Governors)
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Stability: Ability to return to equilibrium after a disturbance. (See 2.5)
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Sensitiveness: Range of speed operation for a given lift. $$\displaystyle S = \frac{N_2 - N_1}{N_m} $$.
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Isochronism: Zero sensitivity; constant speed for all radii.
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Hunting: Continuous fluctuation (oscillation) of speed about mean due to excessive sensitivity.